Question: Let $f$ be a vector-valued function defined by $f(t)=(-2\ln(t+e),6\log_4(4t))$. Find $f$ 's second derivative $f''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac{2}{(t+e)^2},-\dfrac{6}{t^2\ln(4)}\right)$ (Choice B) B $\left(-\dfrac{4}{(t+e)^2},-\dfrac{6}{t^2[\ln(4)]^2}\right)$ (Choice C) C $\left(-2[\ln(t+e)]^3,\dfrac6{16t^2}\right)$ (Choice D) D $\left(-\dfrac{2}{t+e},\dfrac{6}{t\ln(4)}\right)$
Explanation: We are asked to find the second derivative of $f$. This means we need to differentiate $f$ twice. In other words, we differentiate $f$ once to find $f'$, and then differentiate $f'$ (which is a vector-valued function as well) to find $f''$. Recall that $f(t)=(-2\ln(t+e),6\log_4(4t))$. Therefore, $f'(t)=\left(-\dfrac{2}{t+e},\dfrac{6}{t\ln(4)}\right)$. Now let's differentiate $f'(t)=\left(-\dfrac{2}{t+e},\dfrac{6}{t\ln(4)}\right)$ to find $f''$. $f''(t)=\left(\dfrac{2}{(t+e)^2},-\dfrac{6}{t^2\ln(4)}\right)$ In conclusion, $f''(t)=\left(\dfrac{2}{(t+e)^2},-\dfrac{6}{t^2\ln(4)}\right)$.